SHLDR/AC/SC PROC. Flashcards
LABEL THE IMAGE
A. Superior scapular angle
B. Clavicle
C. Acromion process
D. Humeral head
E. Greater tubercle
F. Glenoid cavity
G. Lateral border of the scapula
H. Coracoid Process
LABEL THE IMAGE
A. Superior scapular angle
B. Acromion process
C. Glenoid cavity
D. Scapular body
E. Clavicle
F. Coracoid process
G. Lesser tubercle
LABEL THE IMAGE
A. Acromion
B. Acromial end/extremity
C. Coracoid process
D. Superior scapular angle
E. Vertebral border of scapula
F. Sternal end/extremity
G. Vertebral column
LABEL THE IMAGE
A. Acromion
B. Acromial end/extremity
C. Sternal End/Extremity
D. Superior scapular angle
E. Medial clavicle
LABEL THE IMAGE
A. Acromion
B. Clavicle
C. Coracoid process
D. Scapular spine
E. Superior angle
F. Vertebral border
G. Inferior angle
H. Lateral border
I. Neck
J. Glenoid cavity
LABEL THE IMAGE
A. Head of humerus
B. Greater tubercle
C. Intertubercular groove/sulcus
D. Lesser tubercle
E. Anatomic neck
F. Surgical neck
G. Body of humerus
What type of bone is the clavicle?
Long bone
Lateral end of clavicle:
Articulates with:
Medial end of clavicle:
Articulates with:
- lateral or acromial extremity - art. acromion process
- medial or sternal extremity -art. manubrium of sternum & the first costal cartilage.
Joints of shoulder girdle are classified as:
Movement:
Names of joints:
- synovial joints
- freely movable or diarthrotic.
- Scapulohumeral joint – ball and socket
- Acromioclavicular (AC) joint – gliding (or plane)
- Sternoclavicular (SC) joint – gliding (or plane)
Which joint is only joint that connects shoulder girdle and thorax
SC Joint
AP CLAVICLE
TECHNIQUE:
CR:
RESPIRATION: (& WHY)
ANATOMY:
TECHNIQUE: 70 - 75 kV @ 8 - 12 mAs
CR: PERP. TO MIDCLAVICLE
RESPIRATION: SUSPEND @ END OF INSPIRATION (ELEVATES CLAV)
ANATOMY: Entire clavicle, both AC and SC joints
- Midclavicle is superimposed on the superior scapular angle
AP AXIAL - CLAVICLE
TECHNIQUE:
CR:
ANGLE: (WHY)
RESPIRATION: (& WHY)
ANATOMY:
TECHNIQUE: 70-75 KV @ 8-12 MAS
CR: MID CLAVICLE
ANGLE: 15-30* CEPHALIC (DEGREE OF FX DISPLACEMENT)
RESPIRATION: SUSPEND AT END OF INSPIRATION (ELEVATES CLAV)
ANATOMY: Clavicle is above scapula and the ribs
-Medial end of clavicle is superimposed over 1ST/2ND ribs
LABEL THE IMAGE
A. SC joint
B. Sternal extremity of clavicle
C. Body of clavicle
D. Acromial extremity of clavicle
E. AC joint
F. Scapulohumeral joint
AP AC JOINT
AKA:
ROUTINE:
TECHNIQUE:
SID:
WHY PREFORMED:
IR:
AKA: PEARSON METHOD
ROUTINE: AP BILAT W/O WEIGHTS & AP BILAT W/ WEIGHTS
TECHNIQUE: 75-85 KV @ 16-24
SID: 72 SID
WHY PREFORMED: POSS. AC JOINT SEPERATION
IR: LANDSCAPE / TOP OF IR 2 IN ABOVE SHOULDER
WHAT IS PEARSON METHOD?
WHAT IS DONE PRIOR TO PEASRON METHOD & WHY?
WHAT IS INDICATED IF THERE IS A WIDENING JOINT SPACE?
PEARSON = AP AC JOINTS W & W/O WEIGHTS
- CLAVICLE OR SHOULDER XRAY FIRST R/O FX
- WIDENING JOINT SPACE = SEPERATION OF AC JOINTS
AP AC JOINTS
AKA:
CR:
RESPIRATION:
HOW PREFORMED?
ANATOMY:
AKA: PEARSON METHOD
CR: PERP, BTWN AC JOINTS (1IN ABOVE JUG. NOTCH)
RESPIRATION: SUSPENDED
PREFORMED: 2 EXPOSURES, 1 W. WEIGHT, 1 W/O WEIGHT. STRESS VIEW W. WEIGHT AROUND WRIST
ANATOMY: Both AC joints, entire clavicles, & SC joints
- AC joints on the same horizontal plane.
-No rotation, Symmetric appearance of SC joints
IF UNILATERAL PEARSON - WHAT IS CR?
WHERE ARE WEIGHTS? WHY?
CR: 1IN BELOW AC JOINT OF AFFECTED SIDE
8-10 LBS AROUND WRIST: HOLDING WEIGHTS MAKE SHOULDERS CONTRACT
WHAT PROJECTION?
LEFT IMAGE = PEARSON W/0 WEIGHT
RIGHT IMAGE = PEARSON W. WEIGHTS
PA SC JOINTS
TECHNIQUE:
CR:
RESPIRATION:
ANATOMY:
TECHNIQUE: 75-85 @ 10-12
CR: PERP TO MSP @ T2-T3 (APROX. 3IN DISTAL TO C7)
RESPIRATION: SUSPEND ON EXPIRATION
ANATOMY: BILAT SC JOINS, MEDIAL CLAVICLES THROUGH RIBS & LUNGS
ANTERIOR OBLIQUE POSITIONS SC JOINTS
______ & ________
POSITIONS:
CR:
RESPIRATION:
ANATOMY:
RAO (RIGHT) & LAO (LEFT)
POSITIONS: 10-15* ROTATION
CR: PERP TO T2-T3 (3IN DISTAL TO C7), 1-2IN LATERAL MSP (TOWARD UPSIDE)
RESPIRATION: SUSPEND EXPIRATION
ANATOMY: Sternoclavicular joint on the downside seen
- Manubrium & medial clavicle on downside
- SC joint on upside is foreshortended
HOW DOES CR ENTER ON PA OBLIQUE SC JOINTS?
WHAT ANGLE MAY BE USED ON PA OBLIQUE SCJ?
CR ENTERS UPSIDE, AND VISUALIZES DOWNSIDE
CR: 15* ACROSS PATIENT (UNAFFECTED SIDE TOWARD AFFECTED SIDE)
TWO SURFACES OF SCAPULA:
THREE BORDERS:
THREE ANGLES:
THICKEST PART OF SCAP BODY:
- Costal (anterior) and Dorsal (posterior)
- Lateral (axillary), Medial (vertebral), Superior
- Superior, Inferior, Lateral
- LATERAL ANGLE
WHAT IS CORACOID PROCESS? WHERE LOCATED?
WHERE IS SCAPUHUMERAL JOINT? TYPE OF JOUNT? OTHER NAMES:
- Coracoid process = beaklike process projects anteriorly beneath the clavicle. (located on scapula)
- BETWEEN GLENOID CAVITY & HEAD OF HUMERUS
- SYNOVIAL, BALL&SOCKET
- AKA: GLENOHUMERAL OR SHOULDER JOINT
SHOULDER GIRDLE CONSISTS OF:
PROXIMAL HUMERUS HAS _____ ANTERIORLY & ______LATERALLY, WITH ________ IN BETWEEN.
ACROMION OVERHANGS SHOULDER ________-
- CLAVICLE & SCAPULA
- LESSER TUBERICLE ANT, GREATER TUBERICAL LATERALLY & INTERTUBERCLE GROOVE BTWN
-POSTERIORLY
ROUTINE PROJECTIONS OF SHOULDER:
ADVANCED:
TECHNIQUES:
SID:
R: AP EXTERNAL, INTERNAL & PA OBLIQUE (SCAP Y)
- NUETRAL (IF TRAUMA)
A: LAWRENCE, GRASHEY & NEER METHODS
- 70-75 @ 10-16 MA / SCAP Y: 70-75 @ 20-25
SID: 40 IN
WHICH POSITION IS THE ARM IN THESE XRAYS:
FIRST = EXTERNAL ROTATION
SECOND = INTERNAL ROTATION
THIRD = NEUTRAL
AP SHOULDER - EXTERNAL ROTATION
CR:
ANATOMY:
EPICONDYLES:
CR: PERP. TO 1 IN INFERIOR TO CORACOID PROCESS
EPICONDYLES: PARALLEL WITH IR
ANATOMY: Greater tubercle in full profile laterally.
- Lesser tubercle superimposed over humeral HEADD
-Humeral head in profile medially.
AP SHOULDER - INTERNAL ROTATION
CR:
ANATOMY:
EPICONDYLES:
CR: PERP. TO 1 IN INFERIOR TO CORACOID PROCESS
EPICONDYLES: PERP. TO IR
ANATOMY: LESSER TUBERCLE IN FULL PROFILE MEDIALLY
- GREATER TUBERCLE ANTERIOR, SUPERIMP NY HUMERAL HEAD
- LATERAL VIEW OF PROX HUMERUS
AP SHOULDER - NEUTRAL
CR:
ANATOMY:
EPICONDYLES:
PERP. TO 1 IN INFERIOR TO CORACOID PROCESS
EPICONDYLES: 45* ANGLE WITH IR
ANATOMY: GREATER TUBERCLE IS PARITALLY IN PROFILE LATERALLY, LESSER TUBERCLE ANTERIORLY
IDENTIFY A & B
A. GREATER TUBERCLE
B. LESSER TUBERCLE
PA OBLIQUE SHOULDER
AKA:
POSITION:
CR:
ANATOMY:
SCAPULAR Y
PATIENT 45-60* FROM MCP TO IR
CR: PERP. TO SCAPULOHUMERAL JOINT
ANATOMY: ACROMION & CORACOID PROCESS FORM “Y” ARMS & SCAP BODY FORMS LEG OF “Y”
HOW TO IDENTIFY SHOULDER DISLOCATION
- Anterior dislocation- humeral head sits anteriorly, below the coracoid process.
- Posterior dislocation- humeral head is seen posteriorly, below the acromion.
-
No dislocation or injury: Humeral head and glenoid cavity, and the scapular body and humeral shaft are superimposed
- Humeral head is superimposed over the junction of the “Y”
Pathology in Images
LEFT IMAGE - Posterior Dislocation (humoral head away from body)
RIGHT IMAGE - Anterior Dislocation (numeral head over body)
INFEROSUPERIOR AXIAL PROJECTION
AKA
CR
POSITION
RESPIRATION
ANATOMY
AKA- LAWRENCE METHOD
CR - medially 25-30* horizontal to axilla & humeral head
POSITION - Arm abducted 90* from body, humerus in external position
-Shoulder elevated 2 in
RESPIRATION - suspend
ANATOMY - Open glenohumeral joint space
- Glenoid cavity in profile
- Coracoid process in profile
- Lesser tubercle of humerus in profile
- Acromioclavicular joint
In the Lawrence method, the greater the arm abduction =
What do you do if arm is less than 90* abduction in lawrence method?
Is Lawrence method trauma?
Alternative
- greater arm abd = greater CR angle
- less than 90* = CR 15-20* (decrease)
- Lawrence is nontrauma
Lawrence Method, Alternative position = _____________________
Demonstrates:
What is clements modification? Why done?
- exaggerated external rotation (thumb pointed down & post. 45*)
- Demon. Hill-Sach Defect
-Clements is alternative to Lawrence position. Arm straight up, CR perp to IR (if cant abduct - CR 5-15* perp to axilla)
What is Hills-Sach defect? What projection demonstrates this?
Which scapular structures is most posterior?
A. Coracoid process. B. Glenoid cavity
C. Scapular notch. D. Acromion
- Compression fx of humeral head
- Shown on exaggerated external rotation (lawrence method)
- Acromion
LABEL THE IMAGE
What projection is this?
A. Lesser tubercle
B. Surgical neck
C. Acromion
D. Spine of scapula
E. Coracoid process
F. Head of humerus
G. Glenoid fossa
- Lawrence Method
AP oblique projection – GLENOID CAVITY
AKA:
CR:
POSITION:
RESPIRATION:
ANATOMY:
AKA: Grashey
CR: Perp. to IR @scapulohumeral joint.
POSITION: 35-45* rotation toward affected side (scap. parallel)
RESPIRATION: Suspended
ANATOMY: The scapulohumeral joint is open
- Glenoid cavity in profile w/o superimposition of humeral head
Tangential SUPRASPINATUS OUTLET
AKA:
CR:
ANGLE:
POSITION:
RESPIRATION:
ANATOMY:
AKA: NEEE METHOD
CR: 10-15* caudad @ superior margin of humeral head
POSITION: PA (or face board) 45-60* rotated from MCP, affected side to IR
RESPIRATION:
ANATOMY: Coracoacromial arch for supraspinatus outlet region for possible shoulder impingement.
- Supraspinatus outlet region open, free of superimp. by humeral head
- scapula in lateral profile, “Y” view
Which AP prox. shoulder projections will demonstrate the lesser tubercle in profile medially?
A. External B. Neutral C. Internal
Which demonstrates coracoacromial arch?
a. Scapular Y b. Lawrence c. Neer
d. Grashey
How much is the body rotation for a posterior oblique position (Grashey method)?
- Internal
- NEER
-35-45*
For an inferosuperior axial shoulder projection,
1. The patient’s shoulder is elevated on a sponge or washcloth.
2. The patient’s head is tilted and rotated toward the affected shoulder.
3. The patient’s affected arm is externally rotated.
4. 25 to 30 degree central ray to lateral body surface angle is used if the arm is abducted at a 90-degree angle to the torso.
- 3 & 4
For an AP oblique shoulder projection (Grashey method), the
1. patient’s midcoronal plane is rotated to a 45-degree angle with the IR. 2. central ray is centered to the scapulohumeral joint.
3. patient is rotated toward the affected shoulder.
4. image is obtained with the patient in an upright or supine position.
1, 2, 3 & 4
A referring physician suspects that a subacromial spur may be the cause for a patient’s shoulder impingement. She asks the technologist for a projection that would best demonstrate any possible spurs in the supraspinatus outlet. Which of the following projections would accomplish this objective?
a. Tangential projection with 10° to 15° caudad angle
b. Tangential projection with 10° to 15° cephalad angle
c. AP oblique shoulder with 45° caudad angle
d. AP shoulder with 10° to 15° caudad angle
A. Tangential projection with 10° to 15° caudad angle
Which of the following are clearly demonstrated on the inferosuperior axial projection of the shoulder joint?
Proximal humerus
Scapulohumeral joint Acromioclavicular articulation
1, 2 & 3 (ALL)
The PA oblique projection of the shoulder joint (scapular Y) is performed in which of the following positions?
a. AP. b. Lateral c. RAO or LAO
d. RPO or LPO
RAO & LAO
The glenohumeral joint space is demonstrated as an open space on a(n):
- Inferosuperior axial shoulder.
- Transthoracic lateral shoulder.
- AP oblique shoulder (Grashey).
- PA oblique scapular Y shoulder
Inferosuperior axial shoulder & AP oblique shoulder (Grashey).
